Question

A cannon is fired from a hill \(116.7 \mathrm{~m}\) high at an angle of \(22.7^{\circ}\) with respect to the horizontal. If the muzzle velocity is \(36.1 \mathrm{~m} / \mathrm{s}\), what is the speed of a 4.35 -kg cannonball when it hits the ground \(116.7 \mathrm{~m}\) below?

Short answer

Question: Given that the initial velocity of a cannonball is 36.1 m/s at an angle of 22.7° with respect to the horizontal, and it is fired from a height of 116.7 m, determine the speed of the cannonball when it hits the ground. Answer: To determine the speed of the cannonball, first find the horizontal and vertical components of the initial velocity using trigonometric formulas. Then, calculate the time it takes for the cannonball to hit the ground using a kinematic equation. Finally, calculate the final velocities in both horizontal and vertical directions and determine the magnitude of the final velocity vector, which is the speed of the cannonball when it hits the ground.

Step-by-step solution

Find the horizontal and vertical components of the initial velocity

To find the initial velocity components, we can use the following trigonometric formulas: \\ $$ v_{0x} = v_0 \cos(\theta) $$ $$ v_{0y} = v_0 \sin(\theta) $$ where \(v_0\) is the initial velocity, \(\theta\) is the angle with respect to the horizontal, and \(v_{0x}\) and \(v_{0y}\) are the horizontal and vertical components of the initial velocity, respectively. Using the given values, we get: $$ v_{0x} = 36.1\mathrm{~m/s}\cos(22.7^\circ) $$ $$ v_{0y} = 36.1\mathrm{~m/s}\sin(22.7^\circ) $$

Calculate the time to hit the ground

We can use the kinematic equation to find the time it takes for the cannonball to hit the ground: $$ y = y_0 + v_{0y}t - \frac{1}{2}gt^2 $$ where \(y\) is the final vertical position, \(y_0\) is the initial vertical position, \(v_{0y}\) is the initial vertical velocity, \(t\) is the time, and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)). Since the cannonball hits the ground, \(y = 0\), and because the hill is \(116.7\ \mathrm{m}\) high, \(y_0 = -116.7\mathrm{~m}\). Now, we can solve for the time \(t\) using a quadratic equation.

Find the final velocities and speed

The horizontal velocity remains constant throughout the motion, so the final horizontal velocity is equal to the initial horizontal velocity: $$ v_{fx} = v_{0x} $$ The final vertical velocity can be calculated using another kinematic equation: $$ v_{fy} = v_{0y} - gt $$ The speed of the cannonball when it hits the ground can be found by calculating the magnitude of the final velocity vector, which can be computed using the Pythagorean theorem: $$ v_f =\sqrt{v_{fx}^2 + v_{fy}^2} $$ Plug in the values of \(v_{fx}\) and \(v_{fy}\) obtained from previous steps to find the speed of the cannonball when it hits the ground.

Key concepts

Kinematics

Kinematics is a branch of physics that deals with the motion of objects without considering the forces causing the motion. In projectile motion problems, like the one involving a fired cannonball, we use kinematics to understand how fast and how far the projectile will travel. Knowing the initial conditions, such as the angle of launch and the initial speed, kinematics allows us to track the projectile's path over time.
To solve such problems, we use a set of equations that relate the time of travel, initial velocities, final velocities, and displacements. These equations assume gravity is the only force acting on the projectile (ignoring air resistance), causing a symmetric parabolic trajectory. The kinematic equations help determine important parameters like time of flight, maximum height, and range—all crucial for predicting where and when the projectile will land. In our cannonball scenario, these principles determine the speed at impact, which is the focus of our analysis.

Initial Velocity Components

When dealing with projectile motion, understanding the initial velocity components is crucial. The initial velocity of a projectile, such as the cannonball, can be broken down into two perpendicular components: horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)).

• The horizontal component \(v_{0x}\) is found using\(v_{0x} = v_0 \cos(\theta)\).
• The vertical component \(v_{0y}\) is determined using \(v_{0y} = v_0 \sin(\theta)\).

Here, \(v_0\) is the initial speed of the cannonball, and \(\theta\) is the angle of launch relative to the horizontal plane. This decomposition is vital because these components independently influence the projectile's path. While \(v_{0x}\) governs how far the projectile will travel horizontally, \(v_{0y}\) affects how high and how long the projectile stays in the air. Understanding these components lays the foundation for further calculations, such as finding out how long it takes for the projectile to reach its target.

Final Velocity Calculation

Calculating the final velocity of a projectile when it hits the ground requires knowledge of both its horizontal (\(v_{fx}\)) and vertical (\(v_{fy}\)) velocity components just before impact.

• Horizontal velocity remains constant: \(v_{fx} = v_{0x}\).
• Vertical velocity changes due to gravity: \(v_{fy} = v_{0y} - gt\).

Since gravity constantly accelerates the projectile downwards, \(v_{fy}\) becomes more negative as time goes on. The total impact speed \(v_f\) is the magnitude of the velocity vector, calculated as the square root of the sum of the squares of \(v_{fx}\) and \(v_{fy}\).\[v_f = \sqrt{v_{fx}^2 + v_{fy}^2}\.\]The calculation combines the horizontal and vertical components to find how fast the cannonball is moving right before it lands; this speed reflects the cannonball's hazardous potential upon impact.